cf 1142 C
2024-10-13 08:29:41
忽然觉得这个题很有必要写题解。
移项
y-x^2 = bx+c
那么其实就是找有多少条直线上方没有点
所以就是一个上凸壳有多少条直线/点.
绝妙啊!!!!
太妙了啊!!!!
神乎其技卧槽!!!
(我是傻逼)
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <set>
using namespace std;
typedef long long db;
const double eps=1e-;
const double pi=acos(-);
int sign(db k){
if (k>eps) return ; else if (k<-eps) return -; return ;
}
int cmp(db k1,db k2){return sign(k1-k2);}
struct point{
db x,y;
point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};}
point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};}
point operator * (db k1) const{return (point){x*k1,y*k1};}
point operator / (db k1) const{return (point){x/k1,y/k1};}
bool operator == (const point &k1) const{return cmp(x,k1.x)==&&cmp(y,k1.y)==;}
bool operator <(const point &k1)const {
int c=cmp(x,k1.x);
if(c)return c==-;
return cmp(y,k1.y)==-;
}
};
db cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;}
db dot(point k1,point k2){return k1.x*k2.x+k1.y*k2.y;}
int n;
vector<point> convexHull(vector<point> ps){
int n = ps.size();if(n<=)return ps;
sort(ps.begin(),ps.end());
vector<point> qs(n*);int k=;
for(int i=;i<n;qs[k++]=ps[i++])
while (k>&&cross(qs[k-]-qs[k-],ps[i]-qs[k-])<=)--k;
for(int i=n-,t=k;i>=;qs[k++]=ps[i--])
while (k>t&&cross(qs[k-]-qs[k-],ps[i]-qs[k-])<=)--k;
qs.resize(k-);
return qs;
}
void getUDP(vector<point>A,vector<point>&U,vector<point>&D){
db l=1e7,r=-1e7;
for (int i=;i<A.size();i++) l=min(l,A[i].x),r=max(r,A[i].x);
int wherel,wherer;
for (int i=;i<A.size();i++) if (cmp(A[i].x,l)==) wherel=i;
for (int i=A.size();i;i--) if (cmp(A[i-].x,r)==) wherer=i-;
U.clear(); D.clear(); int now=wherel;
while (){D.push_back(A[now]); if (now==wherer) break; now++; if (now>=A.size()) now=;}
now=wherel;
while (){U.push_back(A[now]); if (now==wherer) break; now--; if (now<) now=A.size()-;}
}
vector<point>p,u,d;
db x,y;
int main(){
scanf("%d",&n);
db mx=;
for(int i=;i<=n;i++){
scanf("%lld%lld",&x,&y);
p.push_back(point{x,y-x*x});
mx+=x;
}
p=convexHull(p);
getUDP(p,u,d);
set<int>s;
for(auto p:u)s.insert(p.x);
printf("%d\n",s.size()-);
}
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